Question

resolva as seguintes equaçoes
a) 3^5x=27
b)5^2x=25
C) 4y+3=1/16

Answer 1

a)

$3^{5x}=27\\ 3^{5x}=3^3\\ 5x=3\\ x=\frac{3}{5}$


b)

$5^{2x}=25\\ 5^{2x}=5^2\\ 2x=2\\ x=\frac{2}{2}\\ x=1$


c)
$4y+3=\frac{1}{16}\\\\ 4y=\frac{1}{16}-3\\\\ 4y=\frac{1-48}{16}\\\\ 4y=\frac{47}{16}\\\\ y=\frac{\frac{47}{16}}{4}\\\\ y=\frac{47}{16}.\frac{1}{4}\\\\ y=\frac{47}{64}$

Mas se a c) for uma equação exponencial, então:

$4^{y+3}=\frac{1}{16}\\\\ 4^{y+3}=4^{-2} \ \ \ \ \ \ (Lembrar \ que: \frac{1}{16}=\frac{1}{4^2}=4^{-2})\\\\ y+3=-2\\ y=-2-3\\ y=-5$